Find matrices complying to given constraints We are given linear mapping of $n$-dimensional vector space, such as:   


*

*It has $n+1$ eigenvectors  

*Any $n$ of them are linearly independent


Find all matrices which could define such a linear mapping.
I'm studying linear algebra on my own, and I have doubts about correctness of my solution. So I would appreciate suggestions or hints about mistakes.
Solution
Given that space is $n$-dimestional, mapping cannot have more than $n$ eigenvalues.
Eigenvectors of different eigenvalues is always linearly independent.
For a given eigenvalue, we could write infinite amount of linearly dependent eigenvectors.  
Now, Let's suppose that matrix, which defines mapping, could be brought to triangular form. than, $det(A - \lambda I) = 0 \to \prod_{i=1}^n (a_{i,i}-\lambda) = 0$.
We will have $n$ independent vectors, if we have $n$ distinct eigenvalues. This will happen for matrix $A$ iff $a_{i,i} \ne a_{j,j} : i,j=1..n$. 
This solution clearly is the best I can do, but it has a suspicious part. Second constraint say "any $n$ of those $n+1$ eigenvectors is linearly independent". But in my solution, we will take for some $\lambda_k$ two vectors from eigenspace, and they would be dependent. So it looks like not 'any'.
Thank you for your help.
 A: Let $S$ be the set with $n+1$ eigenvectors satisfying this property.
Since you have $n$ linear independent eigenvectors in this set, you have a basis for the space, so your linear transformation is diagonalizable. 
Since you have $n+1$ eigenvectors in $S$, you must have at least two eigenvectors associated to the same eigenvalue, let's call this eigenvalue $a$. 
If we remove a eigenvector $v$ associated to $a$ from $S$, we obtain $n$ linear independent eigenvectors, therefore a basis for the space. Thus, $v$ must be a linear combination of the other eigenvectors in $S$ associated to $a$. 
If there exists another eigenvalue, let's say b, if we remove any eigenvector from $S$ associated to $b$, we obtain $n$ linear dependent eigenvectors, because $v$ depends on the others associated to $a$.
So your linear transformation has only one eigenvalue. Since it is diagonalizable, it is a multiple of the identity.
Notice that any multiple of the identity has a set $S$ with this property .
Consider $S=\{e_1,e_2,...,e_n,e_1+e_2+...+e_n\}$, where $\{e_1,...,e_n\}$ is any basis of the space.
